Submission #673709

#TimeUsernameProblemLanguageResultExecution timeMemory
673709vjudge1기지국 (IOI20_stations)C++17
0 / 100
868 ms540 KiB
#include <vector> //#include "IOI20_Stations.h" const int N = 1000; int cnt=0; std::vector<int> labels; /* clarifications: k == n-1 c is sorted Debugging: - Idea => Proof - Code => Accepted */ std::vector<std::vector<int>> g; void dfs(int node, int par) { labels[node] = cnt++; for (int ch : g[node]) { if (ch != par) { dfs(ch, node); } } labels[node] += N * (cnt++); } std::vector<int> label(int n, int k, std::vector<int> u, std::vector<int> v) { /* 1. Construct adjacency list 2. Label every node with its mrg(dt,ft) */ g.clear(); g.resize(n); for (int i = 0; i < n - 1;i++) { g[u[i]].push_back(v[i]); g[v[i]].push_back(u[i]); } cnt = 0; labels.clear(); labels.resize(n); dfs(0, -1); return labels; } /* 1. Now it's easy to find where to countinue the path 2. Tha path will be behined (i.e. the parent of current node), if the distinatin was greater than it 3. The path will be greater than or equal to any of the nighbours 4. If the distination was greater than or equals the node i, and less than any nighbour greater than i, then we need to go there */ int find_next_station(int s, int t, std::vector<int> c) { int sz = c.size(); for (int i = 0; i < sz-1; i++) { // The destination is under this child, Because it's impossible that it was discovered after that, or before that if (t%N >= c[i]%N && t/N <= c[i+2]/N) return c[i]; } return c[sz - 1]; }
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